Solve the following equations for x, y and z:
x + y + z = a; x2 + y2 + z2 = b2; xy = z2
What conditions must a and b satisfy for x, y and z to be distinct positive numbers?
Solution
A routine slog gives z = (a2 - b2)/2a, x and y = (a2 + b2)/4a ± √(10a2b2 - 3a4 - 3b4)/4a.
A little care is needed with the conditions. Clearly x, y, z positive implies a > 0, and then z positive implies |b| < a. The expression under the root must be positive. It helps if you notice that it factorizes as (3a2 - b2)(3b2 - a2). The second factor is positive because |b| < a, so the first factor must also be positive and hence a < √3 |b|. These conditions are also sufficient to ensure that x and y are distinct, but then z must also be distinct because z2 = xy.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
19 Sep 1998
Last corrected/updated 24 Sep 2003